The following sudoku variant was invented by Alexandre Owen Muniz. Here is an own example of this theme.

Fill in the grid so that every row, every column, and every 3x3 box contains the digits 1 through 9. Each number inside the blue cells must be no larger than the number of blue cells in its 3x3 block - the same as the given within that block. Each major diagonal also contains the numbers 1-9. Note that the 9 givens make a magic square.

There a 9 given digits and 9 cages, that is more than 9 clues. The magic sudoku is handmade. But I'd used killer sudoku software (SumuCue and JSudoku) as tools to find this problem. In other cases I make this only by hand or use other programs as tools (Simple Sudoku, Sudoku Susser and an own one). The way to find a puzzle is different for problems with a low number of solution grids and problems with a high number of grids. In both cases it is important to understand how a variant works, which solving techniques are typical. And then there are some guys which publish somewhere variants, help to verify my problems, develop solving methods for special variants or identify my invalid puzzles.

I was able to complete this nice puzzle without "T&E" in a little under a half hour. The two sticking points I found required some careful coloring-type logic on the numbers 1 and 9, but both times considering both the region constraints, the diagonal constraints, and the set of hidden pairs/triplets you've already placed was enough to uniquely place those digits and get several others to follow.

SPARTAN-117 wrote:Has anybody actually managed to do this?? .... Can it be done without T&E?

I had solved it when I made my earlier post. As a relatively unsophisticated solver, I am not averse to T&E (I don't even know what an X-wing is, for example). I used T&E twice. The first was to guess whether the 9 in r8 appeared in r8c3 or r8c7. One of these led (fairly quickly) to a contradiction, the other eventually to the second T&E, which was to guess whether the 1 in the center box appeared in r4c4 or r5c6. Again, one led to a quick contradiction, so bingo.

In both cases, it just so happens I tried the wrong one first, which is always satisfying since I thereby establish uniqueness as well. So, for anybody wondering, I can hereby certify that the solution to this one is indeed unique. (I always worry about this for variants created (partially) by hand.)